An Edge Irregular Reﬂexive k − labeling of Comb Graphs with Additional 2 Pendants

Let G be a connected, simple, and undirrected graph, where V ( G ) is the vertex set and E ( G ) is the edge set. Let k be a natural number. For graph G we deﬁne a total k − labeling ρ such that the vertices of graph G are labeled with { 0 , 2 , 4 , . . . , 2 k v } and the edges of graph G are labeled with { 1 , 2 , 3 , . . . , k e } , where k = max { 2 k v , k e } . Total k − labeling ρ called an edge irregular reﬂexive k − labeling if every two distinct edges of graph G have distinct edge weights, where the edge weight is deﬁned as the sum of the label of that edge and the label of the vertices that are incident to this edge. The minimum k such that G has an edge irregular reﬂexive k − labeling called the reﬂexive edge strength of G. In this paper we determine the reﬂexive edge strength of some comb graphs.


INTRODUCTION
All graphs considered in this paper are finite, connected, and undirected. Let G = (V (G), E(G)) be a simple graph with vertex set V (G) and edge set E(G). Labeling graphs still become an exciting research topic. In 1988, Chartrand et al. [4] defined irregular assignment as a function of the edge set of graph G to the set {1, 2, . . . , k} such that every two distinct vertices have different vertex weights. The vertex weight is defined as the sum of labels of edges that are incident to the vertex. Moreover, Bac'a et.al introduced edge irregular total k−labeling of a graph G as a function from union of edge set and vertex set of graph G to the set {1, 2, . . . , k} such that every two distinct edges have different edge weights. The edge weight is defined as the sum of edge label and all vertex labels that are incident to that edge.

RESULT AND DISCUSSION
2.1. The Comb Graphs with Additional 2 Pendants. The comb graphs with additional 2 pendants, denoted by Comb 2+ n , is defined as the graph with vertex set The comb graphs with additional 2 pendants with n vertices is shown in Figure 1. Moreover, Figure 2 Figure 3 are examples of comb graphs with additional 2 pendants Comb 2+ 6 and Comb 2+ 8 .
The following theorem is about the reflexive edge strength of comb graphs with additional 2 pendants.
n be a comb graph with additional 2 pendants. If n ≥ 3, then Proof. The number of edges of comb graphs with additional 2 pendants Comb 2+ n , for n ≥ 3 is 2n − 3. By Lemma 1.1, the lower bound for res of the comb graphs with additional 2 pendants is as follows, To prove the upper bound we define an edge irregular reflexive k−labeling ρ as follows. All vertices u i for 1 ≤ i ≤ n are labeled with even integers in the following ways: All vertices v i for 2 ≤ i ≤ n − 1 are labeled with even integers in the following ways: All edges u i u i+1 for 1 ≤ i ≤ n − 1 are labeled as follows: All edges u i v i for 2 ≤ i ≤ n − 1 are labeled as follows: The biggest vertex label is ρ v (u n ) = 2 n 3 and the biggest edge labels are ρ e (u n−1 u n ) = 2 n−1 3 − 1 and ρ e (u n−1 v n−1 ) = 2 (n−1)−1 3 = 2 n−2
It is clear that the edge weights of all edges in Comb 2+ n are distinct integers. Thus, the total k−labeling ρ is an edge irregular reflexive k−labeling of Comb 2+ n and k is the reflexive edge strength of Comb 2+ n . Figure 4 is the edge irregular reflexive 8−labeling of a comb graph with 2 additional pendants Comb 2+ 8 .  The double comb graphs with additional 2 pendants with n vertices is shown in Figure  5.  Figure 6 is an example of double comb graph with additional 2 pendants DComb 2+ 8 . The number of edges of double comb graph with additional 2 pendants is 3n − 5 for n ≥ 3. Note that 3n − 5 (mod 6) = 3n + 1 (mod 6), for n ∈ N and 3n (mod 6) = 0, n even, 3, n odd.
Theorem 2.2. Let DComb 2+ n be a double comb graph with additional 2 pendants. If n ≥ 3, then Proof. We know that the graph DComb 2+ n , for n ≥ 3 has 3n − 5 edges. By Lemma 1.1, the lower bound for res of comb graph with additional 2 pendants DComb 2+ n is as follows: The next step, we determine the upper bound of res(DComb 2+ n ) by defining the function ρ. All vertices u i for 1 ≤ i ≤ n are labeled with even integers as follows: All vertices v i for 2 ≤ i ≤ n − 1 are labeled with even integers in the following ways: Furthermore, all vertices u i for 2 ≤ i ≤ n − 1 are labeled with even integers as follows: Moreover, all edges u i u i+1 , for 1 ≤ i ≤ n − 1, are labeled below: All edges u i v i , for 2 ≤ i ≤ n − 1, are labeled in the following ways: All edges u i w i , for 2 ≤ i ≤ n − 1, are labeled as follows: The biggest vertex label is ρ v (v n−1 ) = 2 n−1 2 and the biggest edge label is ρ e (u n−1 u n ) = n. We will prove that res(DComb 2+ n ) = k ≤ 3n−5 3 and we consider two cases.
(i.) For n even. Let n = 2x, for some x ∈ N. The vertex label is and the edge label is Moreover, since 3n − 5 (mod 6) = 3n + 1 (mod 6), we have Thus, we obtained that the vertex label and the edge label always less than 3n−5 3 . (ii.) For n odd.
Let n = 2x + 1, for some x ∈ N. The vertex label is and the edge label is Moreover, since 3n − 5 (mod 6) = 3n + 1 (mod 6), we have Thus, we obtained that the vertex label and the edge label always less than 3n−5
Therefore, the edge weights of graph DComb 2+ n under the labeling ρ are the following: It is already showed that the edge weights of all edges in DComb 2+ n are distinct integers. Thus, the total k−labeling ρ is an edge irregular reflexive k−labeling of DComb 2+ n and k is the reflexive edge strength of DComb 2+ n .  Here are example of some circulant graphs. Figure 1 and Figure 2 are example of circulant graph C 6 (2, 3) and C 8 (1, 3, 4) with vertex set  In this section, we will discuss about reflexive edge strength of circulant graphs. In [2] it is given the following theorem without proof. Here we give the complete proof. 1. If all vertices of graph C 4 (1, 2) are only labeled with 0 or 2, then the edges of graph C 4 (1, 2) can be labeled with at least 1, 2, . . . , 6. 2. If one vertex is labeled with 0 and 3 others are labeled with 2, then there are 3 edges whose endpoints are labeled with 0 and 2. Therefore, all three edges must be labeled with 1, 2, 3. Moreover, there are 3 edges whose endpoints are only labeled with 2. Hence, all three edges also must be labeled with 1, 2, 3. But, there will be 2 edges that have edge weights 5, that is 0 + 3 + 2 and 2 + 1 + 2. 3. If one vertex is labeled with 2 and 3 others are labeled with 0, then there are 3 edges whose endpoints are labeled with 0 and 2. Hence, all three edges must be labeled with 1, 2, 3. Furthermore, there are 3 edges whose endpoints are only labeled with 0. Therefore, all three edges also must be labeled with 1, 2, 3. But, there will be 2 edges that have edge weights 3, that is 0 + 1 + 2 and 0 + 3 + 0. 4. If two vertices are labeled with 2 and 2 others are labeled with 0, then there are 4 edges whose endpoints are labeled with 0 and 2. Consequently, those edges must be labeled with 1, 2, 3, 4.
1. If all vertices of graph C 5 (1, 2) are only labeled with 0 or 2, then the edges of graph C 4 (1, 2) can be labeled with at least 1, 2, . . . , 10. 2. If 4 vertices are labeled with a and one vertex is labeled with 0, for a = b and a, b ∈ {0, 2, 4}, then there are 6 edges that are incident to two vertices labeled with a. Hence, those edges must be labeled with 1, 2, . . . , 6. 3. If two vertices are labeled with a and 3 others are labeled with b, where a and b are distinct elements for a, b ∈ {0, 2, 4}, then there are 6 edges that are incident to vertex labeled with a and b. Therefore, that the six edges must be labeled with 1, 2, . . . , 6. Consequently, for any a, b, and c, it will always be obtained the same edge weights.
Next, we will show that the reflexive edge strength of graphs C 6 (1, 2) is 5. Suppose res(C 5 (1, 2)) = 4. It means that the graph C 6 (1, 2) can be labeled with edge irregular total 4−labeling. Therefore, all the edges of graph C 6 (1, 2) have different edges weights and all the vertices are labeled with even numbers. The number of edges of graph C 6 (1, 2) are 12.
1. If all vertices of graph C 6 (1, 2) are only labeled with 0 or 2, then the edges of graph Consequently, for any a, b, and c, it will always obtained the same edge weights.  Thus, the same edge weights will always be acquired for any a, b and c. Figure 10 is the edge irregular reflexive 10−labeling of a double comb graph with 2 additional pendants Comb 2+ 112 . The following theorem has been proof in [2]. Theorem 3.2. Let C n (1, 2) be a circulant graph. If n ≥ 10, then res(C n (1, 2)) = 2n 3 , n ≡ 0, 2(mod 3) and n ≥ 10, 2n 3 + 1, n ≡ 1(mod 3) and n ≥ 11.